# Math Help - convergent sequences

1. x

2. Originally Posted by britnumber
Let X=(x_n) be a sequence of positive real numbers such that = L >1. Show that X is not a bounded sequence and hence is not convergent
Let $\epsilon > 0$ so that $L - \epsilon > 1$.

And therefore there is $N$ so that if $n\geq N$ then $\left| \frac{x_{n+1}}{x_n} - L \right| < \epsilon$.

Therefore, $L - \epsilon< \frac{x_{n+1}}{x_n} \implies (L - \epsilon)x_n < x_{n+1}$.

Therefore,
$(L - \epsilon)x_N < x_{N+1}$
$(L - \epsilon)x_{N+1} < x_{N+2}$
$(L - \epsilon)x_{N+2} < x_{N+3}$
...
Thus,
$(L - \epsilon)^2x_N < x_{N+2}$
$(L - \epsilon)^3x_N < x_{N+3}$
...
In general,
$(L - \epsilon)^kx_N < x_{N+k}$

But $(L - \epsilon)^k$ can be made arbitary large since $L - \epsilon > 1$.
Therefore, $\{ x_n\}$ is unbounded.